232 ANALYTICAL MECHANICS
TORQUE METHOD. — The cylinder is acted upon by four forces — its weight Mg, the normal reaction N, the frictional reaction F, and the tensile force of the string T. Taking the moments about the element of
contact we obtain
G = T • 2 a,
where a is the radius of the cylinder. The other forces do not have moments about the element of contact. But considering the motion of the falling body we find that
mv = mg — T,
where v is the acceleration of the falling body. Therefore G — m (g — v) • 2 a.
Substituting this value of G in the torque equation, Ico = 2 am (g — v),
where / is the moment of inertia of the cylinder about the element of contact, and co the angular acceleration. But since the highest element of the cylinder has the same linear velocity as the ribbon and the falling body,
we have 2 oco = v, and consequently co = — . Making this substitution
2t CL
in the torque equation
2~ =2am(a-t>),
v =
ENERGY METHOD. — Supposing the initial velocities to be zero and equating the gain in the kinetic energy of the system to the loss in potential energy we have
i 7co2 + J mv* = mgh,
where h is the distance fallen through by the body. Differentiating the last equation with respect to the time,
/coco + mvv = mgh.
But h = V, co = ~- , and cb = ~- . Making these changes and solving for 2 d Z Q,
v we obtain
. __ 4a2m _ A 2am
m m
which are the expressions obtained by the torque method.